## Precalculus (10th Edition)

$\frac{1}{cos\theta}$
The identity says: $tan(\theta)=\frac{sin\theta}{cos\theta}$ and $csc(\theta)=\frac{1}{sin(\theta)}$. Hence $tan(\theta)\cdot csc(\theta)=\frac{sin\theta}{cos\theta}\cdot \frac{1}{sin(\theta)}=\frac{1}{cos\theta}$